Problem: A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.

There are two possible outcomes of a game:

  * the game may result in a tie, then both teams get 1 point; 
  * one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points. 



The score of a team is the number of points it gained during all games that it played.

You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.

Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.

Output

For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.

The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.

All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.

Example

Input


2
2
3


Output


0 
1 -1 1 

Note

In the first test case of the example, both teams get 1 point since the game between them is a tie.

In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).

Solution (in CPP):

Solution:
// Author: Vinay Khilwani
// Language: C++
 
// @vok8: Codeforces, AtCoder, LeetCode, HackerEarth, TopCoder, Google, FB, CSES, Spoj, GitHub
// @vok_8: CodeChef, GFG
// @vok8_khilwani: HackerRank
 
// Never Stop Trying.
// Trying to be Better than Myself.
 
// while(true)
// {
//     if(AC)
//     {
//         break;
//     }
//     else if(Contest Over)
//     {
//         Try.
//         Check out Editorial.
//         Understand.
//         Find out your Mistake.
//         Learn the topic (if new).
//         Solve Problems on that topic (if new).
//         Upsolve that problem.
//         break;
//     }
//     else
//     {
//         Try.
//         Use Pen-Paper.
//         Find errors, edge cases, etc.
//         continue;
//     }
// }
 
// Optimizations
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
#pragma GCC optimize("Os")
 
// Libraries
#include <bits/stdc++.h>
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds;
 
// Debugging
#define dbg(a) cerr<<a<<"\n";
#define debug_a(a) for(auto x:a) {cerr<<x<<" ";} cerr<<"\n";
#define debug_b(a) for(auto x:a) {cerr<<"["<<x.first<<", "<<x.second<<"]"<<"\n";} cerr<<"\n";
#define debug_c(a) for(auto x:a) {debug_a(x)} cerr<<"\n";
#define debug_d(a) for(auto x:a) {debug_b(x)} cerr<<"\n";
#define debug_e(a) cerr<<"["<<a.first<<", "<<a.second<<"]"<<"\n";
 
// Defines
#define fast ios_base::sync_with_stdio(false); cout.tie(NULL); cin.tie(NULL);
#define loop(i,a,n) for(int i=a; i<n; i++)
#define rloop(i,a,n) for(int i=a; i>=n; i--)
#define fr(i,a,n,b) for(int i=a; i<n; i+=b)
#define rfr(i,a,n,b) for(int i=a; i>=n; i-=b)
#define IN cin>>
#define OUT cout<<
#define nl "\n"
#define sz(a) int(a.size())
#define all(a) (a).begin(),(a).end()
#define each(a,b) for(auto &a:b)
#define pb push_back
#define set_bits(a) __builtin_popcountll(a)
#define ar array
#define write(a) for(auto x:a) {OUT x<<" ";} OUT endl;
#define read(a) for(auto &x:a) {IN x;}
// #define oset tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
using ll=long long int;
using ld=long double;
using pll=pair<ll,ll>;
using pii=pair<int,int>;
using vll=vector<ll>;
using vi=vector<int>;
const ll mod=(ll)(1e9)+7LL;
const ll M=998244353LL;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
const ld pi=acos(-1);
 
// General Functions
ll gcd(ll a, ll b)
{
    return (b?gcd(b,a%b):a);
}
ll P(ll B, ll power, ll modulo) //Fast Power
{
    ll ans=1LL;
    while(power>0LL)
    {
        if(power%2LL==1LL)
        {
            ans=(ans*B)%modulo;
        }
        B=(B*B)%modulo;
        power/=2LL;
    }
    return ans;
}
bool isPrime(ll n)
{
    if(n<=1LL)
    {
        return false;
    }
    if(n<=3LL)
    {
        return true;
    }
    if(n%2==0LL || n%3==0LL)
    {
        return false;
    }
    for(ll i=5LL; (i*i)<=n; i+=6LL)
    {
        if(n%i==0LL || n%(i+2LL)==0LL)
        {
            return false;
        }
    }
    return true;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll get_rand(ll l, ll r)
{
    uniform_int_distribution<ll> uid(l,r);
    return uid(rng);
}
void vok()
{
    fast
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
        freopen("error.txt","w",stderr);
    #endif
}
 
// Global Variables
const int mxN=int(1e5)+100;

// Solver Function(s)
void solve()
{
    int n;
    IN n;
    if(n%2)
    {
        loop(i,0,n)
        {
            bool curr=true;
            loop(j,i+1,n)
            {
                if(curr)
                {
                    OUT 1<<" ";
                }
                else
                {
                    OUT -1<<" ";
                }
                curr^=true;
            }
        }
        OUT nl;
    }
    else
    {
        int res[n][n];
        loop(i,0,n)
        {
            loop(j,0,n)
            {
                res[i][j]=-1;
            }
        }
        loop(i,0,n)
        {
            res[n-i-1][i]=0;
        }
        loop(i,0,n)
        {
            bool curr=false;
            if(i%2)
            {
                curr=true;
            }
            loop(j,0,n)
            {
                if(res[i][j]==0)
                {
                    continue;
                }
                if(i==j)
                {
                    continue;
                }
                if(curr)
                {
                    res[i][j]=1;
                }
                curr^=true;
            }
        }
        loop(i,0,n)
        {
            loop(j,i+1,n)
            {
                OUT res[i][j]<<" ";
            }
        }
        OUT nl;
    }
}   
 
// Main Function
int main()
{
    vok();
    int t=1;
    IN t;
    while(t--)
    {
        solve();
    }
    return 0;
}